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netineya
14 days ago
15

Modern commercial airliners are largely made of aluminum, a light and strong metal. But the fact that aluminum is cheap enough t

hat airplanes can be made out of it is a bit of historical luck. Before the discovery of the Hall-Héroult process in , aluminum was as rare and expensive as gold. What would happen if airplanes had to be made of steel? The fuselage of the Airbus A380, which can carry passengers, is approximately a hollow aluminum cylinder without ends, long, wide, and thick (see sketch at right). The fuselage of an airplane Suppose this fuselage was made of steel (density ) instead of aluminum (density ), and let's say the average passenger has a mass of . We'll also assume the engines can't lift any greater mass than they already do. Calculate the number of passengers that the Airbus A380 could carry if its fuselage was made of steel.
Chemistry
1 answer:
lorasvet [956]14 days ago
5 0

Respuesta:

Un avión fabricado con aluminio puede transportar una mayor cantidad de pasajeros comparado con uno de acero.

Explicación:

La masa total que el avión es capaz de levantar es:

m_{tot}=m_{fuselage}+m_{passangers}

Para el aluminio:

m_{tot}=m_{fus-Al}+m_{pas-Al}

m_{fus-Al}=\delta _{Al}*V_{fuselage}

y

V_{fuselage}=\frac{\pi *L}{4}*[D^2-(D-e)^2]

donde:

  • L es longitud
  • D es diámetro
  • e es grosor

m_{tot}=\delta _{Al}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Al}

Para el acero (mismo procedimiento):

m_{tot}=\delta _{Steel}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Steel

Sabiendo que la masa total que el avión puede levantar es constante y que el aluminio tiene una densidad menor que la del acero, podemos afirmar que el avión de aluminio puede levantar un mayor número de pasajeros.

También es posible estimar un peso promedio de los pasajeros para calcular cuántos podría soportar.

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A 8.5-liter sample of a gas has 1.2 mole of the gas. If 0.65 mole of the gas is added, what is the final volume of the gas? Temp
Alekssandra [968]

Answer: 13 liters

Explanation: It is crucial to remember two factors that allow us to carry out this calculation.

Firstly, temperature and pressure must remain unchanged, so these constant values are not considered when calculating volume since they will always be the same.

Secondly, we are dealing with the same gas, with the conditions remaining consistent. Hence, we can proceed.

For 1.2 moles of gas, we have a volume of 8.5 liters. Now, let's determine the volume for 0.65 moles:

0.65 mole * (8.5 liter / 1.2 mole) = 4.25 liters

As we maintain the same gas type, we simply need to total the volumes for each mole amount:

8.5 liters + 4.25 liters = 12.75 liters, which rounds to 13 liters.

7 0
1 day ago
Read 2 more answers
(g) On the graph in part (d) , carefully draw a curve that shows the results of the second titration, in which the student titra
Tems11 [846]

Response:

This is my understanding.

Explanation:

(g) Titration curves

While I can't create two curves on a single graph, I can depict them separately for clarity.

In part (d), the graph indicated an equivalence point at 20 mL.

For the second titration, since the NaOH concentration is doubled, the volume to reach the equivalence point will be halved — 10 mL.

Below are the two titration curves.

(h) Evidence of reaction

Both HCl and NaOH are colorless solutions.

There is no gas released or precipitate formed during their reaction.

It’s likely the student observed that the Erlenmeyer flask heated up, indicating a chemical change.

4 0
16 days ago
A drop of gasoline has a mass of 22 mg and a density of 0.754 g/cm3. What is its volume in cubic centimeters?
Alekssandra [968]
1.22 mg is equivalent to 0.022 grams. Since one gram contains 1000 mg, convert milligrams to grams by dividing by 1000. Calculate volume using mass divided by density: volume equals 0.022 grams divided by 0.754 grams/cm³, resulting in approximately 0.029 or 0.03 cm³. Additionally, weight in newtons equals the mass in kilograms multiplied by gravitational acceleration: weight equals 10 kg times 9.8 m/s², which is 98 newtons. 
5 0
15 days ago
A gold wire has a diameter of 1.00 mm. What length of this wire contains exactly 1.00 mol of gold? (density of Au = 17.0 g/cm3)
KiRa [971]

Answer:

The answer is 7160 cm

Explanation:

Given Data

Diameter = 1 mm

Length =?

Quantity of gold = 1 mol

Density = 17 g/cm³

Steps

1.- Obtain the atomic mass of gold

Atomic mass = 197 g

This means 197g ------------ 1 mol

2.- Find the volume of the wire

Density = mass/volume

Volume = mass/density

Volume = 197/17

Volume = 5.7 cm³

3.- Calculate the wire length

Volume = πr²h

Rearranging for h

h = volume / πr²

Radius = 0.05 cm

Substituting values

h = 5.7/(3.14 x 0.05²)

h = 5.7 / 0.0025

h = 7159.2 cm ≈ 7160 cm

8 0
8 days ago
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If you were given a sample of a cotton ball and a glass stirring rod with identical mass (ex: 5.0 g), which sample would contain
Anarel [852]

Answer:

The sample with more oxygen atoms is a glass stirring rod.

Explanation:

Referring to the periodic table,

the glass stirring rod is composed of Silicon dioxide whereas the cotton ball is made of cellulose.

The molar mass of the glass stirring rod SiO_{2} = 60.08 grams/mole.

The molar mass of the cotton ball C_{6} H_{10}O_{5} = 162.09 grams/mole.

Considering the total number of molecules for oxygen is 32,

therefore,

In the glass stirring rod,

oxygen atoms contained in 5g = \frac{32}{60.07} \times 5

= 2.66 g of oxygen

= \frac{1}{16} \times 2.66

= 0.16625 moles

= 0.16625 x 6.023\times 10^{23}

= 1.001 \times10^{23} atoms

In the cotton ball,

oxygen atoms contained in 5g = \frac{80}{162.09} \times 5

= 2.467 g of oxygen

= \frac{1}{16} \times 2.467

= 0.15418 moles

= 0.15418 \times 6.023 \times 10^{23}

= 0.928 \\ \times10^{23}

Thus, the glass stirring rod has a greater quantity of oxygen atoms.

4 0
9 days ago
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